1900. The Earliest and Latest Rounds Where Players Compete

Problem Description

You have a tournament with n players standing in a row, numbered from 1 to n based on their initial positions (player 1 is first, player 2 is second, etc.).

The tournament consists of multiple rounds, and in each round:

The i-th player from the front competes against the i-th player from the end
Winners advance to the next round
If there's an odd number of players, the middle player automatically advances

For example, with players [1, 2, 4, 6, 7]:

Player 1 competes against player 7
Player 2 competes against player 6
Player 4 (middle) automatically advances

After each round, winners are re-arranged in their original numerical order (ascending) before the next round begins.

Two special players, firstPlayer and secondPlayer, are the best in the tournament - they will always win against any other player except each other. For all other matches between regular players, you can choose who wins.

Given n (total players), firstPlayer, and secondPlayer, you need to find:

The earliest possible round where these two players could compete
The latest possible round where these two players could compete

Return these two values as an array [earliest_round, latest_round].

The key insight is that by controlling the outcomes of matches between regular players, you can influence when the two special players eventually meet, either forcing them to meet as soon as possible or delaying their meeting as long as possible.

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Intuition

The key observation is that we need to track the positions of our two special players as they advance through rounds. Since they always win their matches (except against each other), we can control when they meet by strategically choosing winners in other matches.

Think about what happens in each round:

Players pair up from opposite ends of the row
Winners maintain their original ordering for the next round
The two special players will definitely advance until they meet

The challenge is that after each round, the number of players reduces and their relative positions change. If we have n players in a round, roughly n/2 advance to the next round (plus one more if n is odd).

To find both earliest and latest meeting rounds, we need to explore all possible outcomes of matches between regular players. For each round, we can represent match outcomes as a binary choice - for each pair of regular players, either the front player wins (bit = 1) or the back player wins (bit = 0).

This leads us to a recursive approach with memoization:

State: positions of our two special players (l and r) among n current players
Base case: when l + r = n - 1, they're paired against each other (one from front, one from back)
Transition: try all possible match outcomes for regular players, track where our special players end up in the next round

By using binary enumeration (1 << m combinations for m pairs), we can systematically explore all possible tournament progressions. For each possibility:

Determine which players advance based on our choice
Find the new positions of our special players among winners
Recursively solve for the smaller tournament
Track minimum rounds (earliest) and maximum rounds (latest)

The memoization with @cache prevents recalculating the same state multiple times, as the same configuration of players can be reached through different paths.

Learn more about Memoization and Dynamic Programming patterns.

Solution Approach

The solution uses memoization with binary enumeration to explore all possible tournament outcomes.

Core Function: dfs(l, r, n)

This function returns [earliest_round, latest_round] for players at positions l and r among n total players.

Base Case:

When l + r = n - 1, the two players face each other (one from position l from front, other from position n-1-r from back, which equals l)
Return [1, 1] since they meet in the current round

Recursive Case:

Calculate number of matches: m = n >> 1 (equivalent to n // 2)
- This gives us the number of player pairs competing
Binary enumeration of match outcomes: Loop through i from 0 to 2^m - 1
- Each bit in i represents one match outcome
- If bit j is 1: front player j wins
- If bit j is 0: back player n-1-j wins

Determine winners for each configuration:

win = [False] * n
for j in range(m):
    if i >> j & 1:  # Check if j-th bit is 1
        win[j] = True
    else:
        win[n - 1 - j] = True

If n is odd, middle player automatically advances: win[m] = True
Force our special players to win: win[l] = True and win[r] = True
Their opponents must lose: win[n - 1 - l] = False and win[n - 1 - r] = False

Calculate new positions in next round:

a = b = c = 0
for j in range(n):
    if j == l: a = c  # New position of first special player
    if j == r: b = c  # New position of second special player
    if win[j]: c += 1  # Count winners before current position

Variable c counts winners up to position j
When we reach special player positions, record their new positions among winners

Recursive call and update results:

x, y = dfs(a, b, c)  # Get results for next round
res[0] = min(res[0], x + 1)  # Update earliest (minimum)
res[1] = max(res[1], y + 1)  # Update latest (maximum)

Add 1 to account for the current round

Memoization:

The @cache decorator automatically memoizes results
Prevents recalculating the same state (l, r, n) multiple times

Main Function:

Converts 1-indexed player numbers to 0-indexed positions
Calls dfs(firstPlayer - 1, secondPlayer - 1, n)

The algorithm explores all possible tournament trees where the two special players advance, finding both the shortest and longest paths to their eventual meeting.

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Example Walkthrough

Let's walk through a small example with n = 5, firstPlayer = 1, secondPlayer = 5.

Initial setup: Players [1, 2, 3, 4, 5] where players 1 and 5 are special (always win).

Round 1 Analysis:

We call dfs(0, 4, 5) (positions 0 and 4 for players 1 and 5)
Check base case: 0 + 4 = 4 = 5 - 1, so they face each other now!
Player 1 (position 0) faces Player 5 (position 4 from back, which is position 0 from end)
Return [1, 1] - they meet in round 1

Now let's try n = 7, firstPlayer = 2, secondPlayer = 6:

Initial: Players [1, 2, 3, 4, 5, 6, 7] where players 2 and 6 are special.

Round 1:

Call dfs(1, 5, 7) (positions 1 and 5)
Base case check: 1 + 5 = 6 ≠ 7 - 1, so continue
Number of matches: m = 7 >> 1 = 3
Player 4 (middle) auto-advances

We explore different match outcomes using binary enumeration (8 possibilities):

Example Configuration (i = 3, binary: 011):

Bit 0 = 1: Player 1 beats Player 7
Bit 1 = 1: Player 2 beats Player 6 (Player 2 must win anyway)
Bit 2 = 0: Player 5 beats Player 3
Player 4 auto-advances

Winners: [1, 2, 4, 5] → Player 2 is at position 1, Player 6 didn't advance (eliminated)

This path is invalid since Player 6 must win.

Valid Configuration (i = 2, binary: 010):

Bit 0 = 0: Player 7 beats Player 1
Bit 1 = 1: Player 2 beats Player 6 (forced - Player 2 must win)
Bit 2 = 0: Player 5 beats Player 3
Player 4 auto-advances

But wait - Player 6 at position 5 faces Player 2 at position 1, so Player 6 must also win! Let's correct this:

Valid Configuration for Earliest Meeting: Choose outcomes to bring players 2 and 6 closer:

Player 2 (pos 1) vs Player 6 (pos 5) - they don't face each other yet
Player 1 (pos 0) vs Player 7 (pos 6) - let Player 1 win
Player 2 (pos 1) vs Player 6 (pos 5) - NO, Player 2 faces Player 6's opponent
Actually: Player 2 (pos 1) faces Player 6 (pos 5) from back = position 1 from end

Let me recalculate the pairings:

Position 0 vs Position 6 (Player 1 vs Player 7)
Position 1 vs Position 5 (Player 2 vs Player 6)
Position 2 vs Position 4 (Player 3 vs Player 5)
Position 3 (Player 4) auto-advances

Wait, 1 + 5 = 6 = 7 - 1, so Players 2 and 6 DO face each other in round 1!

Let's try n = 6, firstPlayer = 2, secondPlayer = 5:

Round 1:

Call dfs(1, 4, 6)
Base case: 1 + 4 = 5 ≠ 6 - 1, continue
Matches: Player 1 vs 6, Player 2 vs 5, Player 3 vs 4
Player 2 faces Player 5! Check: position 1 from front vs position 4 from back
Position 4 from back is position 6-1-4 = 1 from end... that's position 4!
So Player at position 1 faces player at position 4

Actually, the pairing works as: position i faces position n-1-i.

Position 1 faces position 6-1-1 = 4 ✓

So players 2 and 5 meet in round 1.

For a case where they don't meet immediately, consider n = 7, firstPlayer = 1, secondPlayer = 4:

Round 1:

Call dfs(0, 3, 7)
Base case: 0 + 3 = 3 ≠ 6, continue
Pairings: 1 vs 7, 2 vs 6, 3 vs 5, 4 auto-advances
Players 1 and 4 don't face each other

By controlling who wins between 2 vs 6 and 3 vs 5, we can influence when players 1 and 4 eventually meet. The algorithm explores all such possibilities to find both the earliest and latest rounds where they can meet.

Solution Implementation

1from functools import cache
2from typing import List
3
4class Solution:
5    def earliestAndLatest(
6        self, n: int, firstPlayer: int, secondPlayer: int
7    ) -> List[int]:
8        """
9        Find the earliest and latest rounds when two players can meet in a tournament.
10      
11        Args:
12            n: Total number of players
13            firstPlayer: Position of the first player (1-indexed)
14            secondPlayer: Position of the second player (1-indexed)
15      
16        Returns:
17            List containing [earliest_round, latest_round]
18        """
19      
20        @cache
21        def find_min_max_rounds(left_pos: int, right_pos: int, remaining_players: int) -> List[int]:
22            """
23            Recursively find minimum and maximum rounds for two players to meet.
24          
25            Args:
26                left_pos: 0-indexed position of left player
27                right_pos: 0-indexed position of right player
28                remaining_players: Number of players remaining in current round
29          
30            Returns:
31                List containing [min_rounds, max_rounds]
32            """
33            # Base case: if players are adjacent in the pairing, they meet next round
34            if left_pos + right_pos == remaining_players - 1:
35                return [1, 1]
36          
37            # Initialize result with worst case bounds
38            min_rounds = float('inf')
39            max_rounds = float('-inf')
40          
41            # Calculate number of matches in this round
42            num_matches = remaining_players >> 1  # Integer division by 2
43          
44            # Try all possible match outcomes (2^num_matches possibilities)
45            for outcome_mask in range(1 << num_matches):
46                # Create array to track which players win their matches
47                winners = [False] * remaining_players
48              
49                # Determine winners based on outcome_mask bits
50                for match_idx in range(num_matches):
51                    if outcome_mask >> match_idx & 1:
52                        # Player at position match_idx wins
53                        winners[match_idx] = True
54                    else:
55                        # Player at symmetric position wins
56                        winners[remaining_players - 1 - match_idx] = True
57              
58                # If odd number of players, middle player gets a bye
59                if remaining_players & 1:
60                    winners[num_matches] = True
61              
62                # Force our two tracked players to win their matches
63                winners[remaining_players - 1 - left_pos] = False
64                winners[remaining_players - 1 - right_pos] = False
65                winners[left_pos] = True
66                winners[right_pos] = True
67              
68                # Calculate new positions after this round
69                new_left_pos = 0
70                new_right_pos = 0
71                position_counter = 0
72              
73                for player_idx in range(remaining_players):
74                    if player_idx == left_pos:
75                        new_left_pos = position_counter
76                    if player_idx == right_pos:
77                        new_right_pos = position_counter
78                    if winners[player_idx]:
79                        position_counter += 1
80              
81                # Recursively find min/max rounds for next round
82                sub_min, sub_max = find_min_max_rounds(
83                    new_left_pos, new_right_pos, position_counter
84                )
85              
86                # Update overall min and max
87                min_rounds = min(min_rounds, sub_min + 1)
88                max_rounds = max(max_rounds, sub_max + 1)
89          
90            return [min_rounds, max_rounds]
91      
92        # Convert to 0-indexed positions and start recursion
93        return find_min_max_rounds(firstPlayer - 1, secondPlayer - 1, n)
94

1class Solution {
2    // Memoization cache: [leftPosition][rightPosition][totalPlayers]
3    // Stores encoded min/max rounds as a single integer
4    static int[][][] memo = new int[30][30][31];
5
6    /**
7     * Find the earliest and latest round where firstPlayer and secondPlayer meet
8     * @param n Total number of players
9     * @param firstPlayer Position of first player (1-indexed)
10     * @param secondPlayer Position of second player (1-indexed)
11     * @return Array of [earliest round, latest round]
12     */
13    public int[] earliestAndLatest(int n, int firstPlayer, int secondPlayer) {
14        // Convert to 0-indexed positions
15        return dfs(firstPlayer - 1, secondPlayer - 1, n);
16    }
17
18    /**
19     * Recursive DFS to find min and max rounds for two players to meet
20     * @param leftPos Position of the left player (0-indexed)
21     * @param rightPos Position of the right player (0-indexed)
22     * @param totalPlayers Total number of players in current round
23     * @return Array of [min rounds, max rounds]
24     */
25    private int[] dfs(int leftPos, int rightPos, int totalPlayers) {
26        // Check memoization cache
27        if (memo[leftPos][rightPos][totalPlayers] != 0) {
28            return decode(memo[leftPos][rightPos][totalPlayers]);
29        }
30      
31        // Base case: players meet when their positions sum to totalPlayers - 1
32        // (they are paired in the tournament bracket)
33        if (leftPos + rightPos == totalPlayers - 1) {
34            memo[leftPos][rightPos][totalPlayers] = encode(1, 1);
35            return new int[] {1, 1};
36        }
37      
38        int minRounds = Integer.MAX_VALUE;
39        int maxRounds = Integer.MIN_VALUE;
40        int halfPlayers = totalPlayers >> 1;  // Number of pairs in current round
41      
42        // Try all possible match outcomes (2^halfPlayers possibilities)
43        for (int matchOutcomes = 0; matchOutcomes < (1 << halfPlayers); matchOutcomes++) {
44            // Track which players win their matches
45            boolean[] isWinner = new boolean[totalPlayers];
46          
47            // Determine winners for each pair based on current matchOutcomes bitmask
48            for (int pairIndex = 0; pairIndex < halfPlayers; pairIndex++) {
49                if (((matchOutcomes >> pairIndex) & 1) == 1) {
50                    // Player at position pairIndex wins
51                    isWinner[pairIndex] = true;
52                } else {
53                    // Player at opposite position wins
54                    isWinner[totalPlayers - 1 - pairIndex] = true;
55                }
56            }
57          
58            // Middle player always advances if odd number of players
59            if ((totalPlayers & 1) == 1) {
60                isWinner[halfPlayers] = true;
61            }
62          
63            // Force our two tracked players to win their matches
64            isWinner[totalPlayers - 1 - leftPos] = false;   // Opponent of left player loses
65            isWinner[totalPlayers - 1 - rightPos] = false;  // Opponent of right player loses
66            isWinner[leftPos] = true;                       // Left player wins
67            isWinner[rightPos] = true;                      // Right player wins
68          
69            // Calculate new positions for tracked players in next round
70            int newLeftPos = 0;
71            int newRightPos = 0;
72            int nextRoundPosition = 0;
73          
74            for (int currentPos = 0; currentPos < totalPlayers; currentPos++) {
75                if (currentPos == leftPos) {
76                    newLeftPos = nextRoundPosition;
77                }
78                if (currentPos == rightPos) {
79                    newRightPos = nextRoundPosition;
80                }
81                if (isWinner[currentPos]) {
82                    nextRoundPosition++;
83                }
84            }
85          
86            // Recursively find min/max rounds for next round configuration
87            int[] result = dfs(newLeftPos, newRightPos, nextRoundPosition);
88            minRounds = Math.min(minRounds, result[0] + 1);
89            maxRounds = Math.max(maxRounds, result[1] + 1);
90        }
91      
92        // Cache and return result
93        memo[leftPos][rightPos][totalPlayers] = encode(minRounds, maxRounds);
94        return new int[] {minRounds, maxRounds};
95    }
96
97    /**
98     * Encode two integers into a single integer for memoization
99     * @param minValue Minimum rounds (stored in upper bits)
100     * @param maxValue Maximum rounds (stored in lower 8 bits)
101     * @return Encoded integer value
102     */
103    private int encode(int minValue, int maxValue) {
104        return (minValue << 8) | maxValue;
105    }
106
107    /**
108     * Decode a single integer back into min and max values
109     * @param encodedValue The encoded integer
110     * @return Array of [min rounds, max rounds]
111     */
112    private int[] decode(int encodedValue) {
113        return new int[] {encodedValue >> 8, encodedValue & 255};
114    }
115}
116

1class Solution {
2private:
3    // Memoization table: dp[leftPosition][rightPosition][totalPlayers]
4    // Stores encoded min/max values for each state
5    int dp[30][30][31];
6  
7public:
8    vector<int> earliestAndLatest(int n, int firstPlayer, int secondPlayer) {
9        // Convert to 0-indexed positions
10        return findMinMaxRounds(firstPlayer - 1, secondPlayer - 1, n);
11    }
12
13private:
14    vector<int> findMinMaxRounds(int leftPlayerPos, int rightPlayerPos, int totalPlayers) {
15        // Check memoization cache
16        if (dp[leftPlayerPos][rightPlayerPos][totalPlayers] != 0) {
17            return decodeMinMax(dp[leftPlayerPos][rightPlayerPos][totalPlayers]);
18        }
19      
20        // Base case: players face each other when their positions sum to totalPlayers - 1
21        // (they are paired in the current round)
22        if (leftPlayerPos + rightPlayerPos == totalPlayers - 1) {
23            dp[leftPlayerPos][rightPlayerPos][totalPlayers] = encodeMinMax(1, 1);
24            return {1, 1};
25        }
26      
27        int minRounds = INT_MAX;
28        int maxRounds = INT_MIN;
29        int halfPlayers = totalPlayers >> 1;  // Number of matches in current round
30      
31        // Try all possible match outcomes (2^halfPlayers possibilities)
32        for (int matchOutcomeMask = 0; matchOutcomeMask < (1 << halfPlayers); matchOutcomeMask++) {
33            // Track which players advance to next round
34            vector<bool> advances(totalPlayers, false);
35          
36            // For each match, determine winner based on current mask
37            for (int matchIdx = 0; matchIdx < halfPlayers; matchIdx++) {
38                if ((matchOutcomeMask >> matchIdx) & 1) {
39                    // Player at position matchIdx wins
40                    advances[matchIdx] = true;
41                } else {
42                    // Player at opposite position wins
43                    advances[totalPlayers - 1 - matchIdx] = true;
44                }
45            }
46          
47            // Middle player always advances if odd number of players
48            if (totalPlayers & 1) {
49                advances[halfPlayers] = true;
50            }
51          
52            // Ensure our two tracked players always advance
53            advances[totalPlayers - 1 - leftPlayerPos] = false;  // Opponent of left player loses
54            advances[totalPlayers - 1 - rightPlayerPos] = false; // Opponent of right player loses
55            advances[leftPlayerPos] = true;   // Left player wins
56            advances[rightPlayerPos] = true;  // Right player wins
57          
58            // Calculate new positions of our players in the next round
59            int newLeftPos = 0;
60            int newRightPos = 0;
61            int nextRoundPosition = 0;
62          
63            for (int playerIdx = 0; playerIdx < totalPlayers; playerIdx++) {
64                if (playerIdx == leftPlayerPos) {
65                    newLeftPos = nextRoundPosition;
66                }
67                if (playerIdx == rightPlayerPos) {
68                    newRightPos = nextRoundPosition;
69                }
70                if (advances[playerIdx]) {
71                    nextRoundPosition++;
72                }
73            }
74          
75            // Recursively find min/max rounds for next state
76            vector<int> nextRoundResult = findMinMaxRounds(newLeftPos, newRightPos, nextRoundPosition);
77            minRounds = std::min(minRounds, nextRoundResult[0] + 1);
78            maxRounds = std::max(maxRounds, nextRoundResult[1] + 1);
79        }
80      
81        // Cache and return result
82        dp[leftPlayerPos][rightPlayerPos][totalPlayers] = encodeMinMax(minRounds, maxRounds);
83        return {minRounds, maxRounds};
84    }
85  
86    // Encode min and max values into a single integer for memoization
87    int encodeMinMax(int minVal, int maxVal) {
88        return (minVal << 8) | maxVal;
89    }
90  
91    // Decode the encoded integer back to min and max values
92    vector<int> decodeMinMax(int encodedVal) {
93        return {encodedVal >> 8, encodedVal & 255};
94    }
95};
96

1// Global memoization table for dynamic programming
2// Stores encoded results for [leftPosition][rightPosition][remainingPlayers]
3const memoTable: number[][][] = Array.from({ length: 30 }, () =>
4    Array.from({ length: 30 }, () => Array(31).fill(0)),
5);
6
7/**
8 * Finds the earliest and latest rounds when two players might compete
9 * @param n - Total number of players
10 * @param firstPlayer - Position of first player (1-indexed)
11 * @param secondPlayer - Position of second player (1-indexed)
12 * @returns Array containing [earliestRound, latestRound]
13 */
14function earliestAndLatest(n: number, firstPlayer: number, secondPlayer: number): number[] {
15    // Convert to 0-indexed positions and start DFS
16    return calculateRounds(firstPlayer - 1, secondPlayer - 1, n);
17}
18
19/**
20 * Recursively calculates the minimum and maximum rounds for two players to meet
21 * @param leftPlayerPos - Position of the left player (0-indexed)
22 * @param rightPlayerPos - Position of the right player (0-indexed)
23 * @param totalPlayers - Total number of players remaining
24 * @returns Array containing [minRounds, maxRounds]
25 */
26function calculateRounds(leftPlayerPos: number, rightPlayerPos: number, totalPlayers: number): number[] {
27    // Check memoization table for cached result
28    if (memoTable[leftPlayerPos][rightPlayerPos][totalPlayers] !== 0) {
29        return decodeResult(memoTable[leftPlayerPos][rightPlayerPos][totalPlayers]);
30    }
31  
32    // Base case: players are paired in this round (symmetric positions)
33    if (leftPlayerPos + rightPlayerPos === totalPlayers - 1) {
34        memoTable[leftPlayerPos][rightPlayerPos][totalPlayers] = encodeResult(1, 1);
35        return [1, 1];
36    }
37
38    let minRounds = Number.MAX_SAFE_INTEGER;
39    let maxRounds = Number.MIN_SAFE_INTEGER;
40    const halfPlayers = totalPlayers >> 1; // Number of pairs in current round
41
42    // Iterate through all possible match outcomes using bitmask
43    for (let matchOutcomes = 0; matchOutcomes < (1 << halfPlayers); matchOutcomes++) {
44        // Array to track which players win their matches
45        const winnerStatus: boolean[] = Array(totalPlayers).fill(false);
46      
47        // Set winners based on current bitmask configuration
48        for (let pairIndex = 0; pairIndex < halfPlayers; pairIndex++) {
49            if ((matchOutcomes >> pairIndex) & 1) {
50                // Player at position pairIndex wins
51                winnerStatus[pairIndex] = true;
52            } else {
53                // Player at symmetric position wins
54                winnerStatus[totalPlayers - 1 - pairIndex] = true;
55            }
56        }
57
58        // Middle player always advances if odd number of players
59        if (totalPlayers & 1) {
60            winnerStatus[halfPlayers] = true;
61        }
62
63        // Ensure our two tracked players always win their matches
64        winnerStatus[totalPlayers - 1 - leftPlayerPos] = false;  // Opponent of left player loses
65        winnerStatus[totalPlayers - 1 - rightPlayerPos] = false; // Opponent of right player loses
66        winnerStatus[leftPlayerPos] = true;  // Left player wins
67        winnerStatus[rightPlayerPos] = true; // Right player wins
68
69        // Calculate new positions for the two players in next round
70        let newLeftPos = 0;
71        let newRightPos = 0;
72        let winnerCount = 0;
73      
74        for (let position = 0; position < totalPlayers; position++) {
75            if (position === leftPlayerPos) {
76                newLeftPos = winnerCount;
77            }
78            if (position === rightPlayerPos) {
79                newRightPos = winnerCount;
80            }
81            if (winnerStatus[position]) {
82                winnerCount++;
83            }
84        }
85
86        // Recursively calculate rounds for next configuration
87        const nextRoundResult = calculateRounds(newLeftPos, newRightPos, winnerCount);
88        minRounds = Math.min(minRounds, nextRoundResult[0] + 1);
89        maxRounds = Math.max(maxRounds, nextRoundResult[1] + 1);
90    }
91
92    // Cache and return the result
93    memoTable[leftPlayerPos][rightPlayerPos][totalPlayers] = encodeResult(minRounds, maxRounds);
94    return [minRounds, maxRounds];
95}
96
97/**
98 * Encodes two numbers into a single integer for storage
99 * @param minValue - Minimum value (stored in higher bits)
100 * @param maxValue - Maximum value (stored in lower bits)
101 * @returns Encoded integer value
102 */
103function encodeResult(minValue: number, maxValue: number): number {
104    return (minValue << 8) | maxValue;
105}
106
107/**
108 * Decodes a single integer back into two numbers
109 * @param encodedValue - The encoded integer value
110 * @returns Array containing [minValue, maxValue]
111 */
112function decodeResult(encodedValue: number): number[] {
113    return [encodedValue >> 8, encodedValue & 255];
114}
115

Time and Space Complexity

Time Complexity: O(n^3 * 2^(n/2))

The analysis breaks down as follows:

The function dfs has at most O(n^2) unique states since parameters l and r can each range from 0 to n-1, and n decreases with each recursive call
For each state, the function iterates through 2^(n/2) possibilities (the for loop with range(1 << m) where m = n >> 1)
Inside this loop, there's an O(n) operation to process the win array and calculate values a, b, and c
Due to memoization with @cache, each unique state is computed only once
The recursion depth is at most O(log n) since n roughly halves with each recursive level
Total complexity: O(n^2) * O(2^(n/2)) * O(n) = O(n^3 * 2^(n/2))

Space Complexity: O(n^3)

The space analysis includes:

The memoization cache stores at most O(n^3) states (considering all possible combinations of l, r, and n)
The recursion stack depth is O(log n)
Each recursive call uses O(n) space for the win array
Total space: O(n^3) for the cache dominates, giving us O(n^3) overall space complexity

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Winner Assignment for Symmetric Matches

A critical pitfall is incorrectly handling the constraint that when two players compete, exactly one must win and one must lose. When forcing the special players to win, you must ensure their opponents lose.

Incorrect approach:

# Only setting winners without considering opponents
winners[left_pos] = True
winners[right_pos] = True

Correct approach:

# First set opponents to lose, then set special players to win
winners[remaining_players - 1 - left_pos] = False
winners[remaining_players - 1 - right_pos] = False
winners[left_pos] = True
winners[right_pos] = True

2. Edge Case: Special Players Competing Against Each Other

When the two special players face each other in the current round (i.e., left_pos + right_pos == remaining_players - 1), trying to force both to win creates a logical contradiction.

Solution: Check this condition first as the base case and return [1, 1] immediately, avoiding the winner assignment logic entirely.

3. Bit Manipulation Confusion

Understanding how the outcome mask works can be tricky. Each bit represents a match outcome, not a player position.

Common mistake:

# Thinking bit i represents player i
if outcome_mask & (1 << player_idx):  # Wrong!

Correct understanding:

# Bit j represents match j (between player j and player n-1-j)
if outcome_mask >> match_idx & 1:
    winners[match_idx] = True
else:
    winners[remaining_players - 1 - match_idx] = True

4. Position Tracking After Elimination

After each round, players are re-ordered based on their original numbers (1 to n), not their current positions. The new position depends on how many winners appear before them in the original ordering.

Incorrect:

# Assuming positions are reassigned sequentially
new_positions = list(range(num_winners))  # Wrong!

Correct:

position_counter = 0
for player_idx in range(remaining_players):
    if player_idx == left_pos:
        new_left_pos = position_counter
    if player_idx == right_pos:
        new_right_pos = position_counter
    if winners[player_idx]:
        position_counter += 1

5. Initialization Bounds for Min/Max

Using incorrect initial values can lead to wrong results.

Pitfall:

min_rounds = 0  # Too small
max_rounds = n   # Might be too large or too small

Solution:

min_rounds = float('inf')
max_rounds = float('-inf')

These ensure any valid value will properly update the bounds during iteration.

Discover Your Strengths and Weaknesses: Take Our 5-Minute Quiz to Tailor Your Study Plan:

1900. The Earliest and Latest Rounds Where Players Compete

Problem Description

Intuition

Solution Approach

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Example Walkthrough

Solution Implementation

Time and Space Complexity

Common Pitfalls

1. Incorrect Winner Assignment for Symmetric Matches

2. Edge Case: Special Players Competing Against Each Other

3. Bit Manipulation Confusion

4. Position Tracking After Elimination

5. Initialization Bounds for Min/Max

Recommended Readings